Question

A geochemist in the field takes a 34.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 22.0 °C, and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to 750.0 mL. Then he filters it and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries, and weighs the crystals. They weigh 0.31 g.

Using only the information above, can you calculate yes the solubility of X in the water at 17.0 °C? If yes, calculate it. Be sure your answer has a unit symbol and 3 significant digits.

Answer 1

The solubility is
`S = 0.00912 \ g/mL`

Step-by-step explanation:

From the question we are told that

The original volume of sample is
`V_o = 34.0 mL = 34 *10^(-3) \ L`

The temperature is
`T = 22.0 ^oC`

The new volume of sample is
`V_n = 750.0 mL = 750 *10^(-3) \ L`

The weight of the crystal is
`X = 0.31 \ g = 0.31 *10^(-3) \ kg`

Now looking at the question we see that 34.0 mL of the sample is saturated with 0.31g of the crystal X

Generally the solubility of X in the water sample at
`22.0 ^oC` can be mathematically evaluate as

`S = (0.31 )/(34.0)`

`S = 0.00912 \ g/mL`